The amazing maze II: searching the maze

This post I will talk about how to solve a maze. If you are interested in how to build a maze. Take a look at my last post.

There are couple ways to solve a maze, the easiest ones are using DFS and BFS. Here, I implemented DFS using recursion, DFS using a stack, and BFS.

package mazeDFS;
import java.util.*;
public class SolveMaze {
 
 List path;
 private Maze m;
 
 public SolveMaze(Maze m) {
  path = new ArrayList ();
  this.m = m;
 }
 /**
  * Using DFS to solve the maze
  */
 public void solveByDFS() {
  boolean[] visited = new boolean[m.grid.length];
  //Stack stack = new Stack ();
  dfs(path, visited, 0);
 }
 //using recursion
 private void dfs(List path, boolean[] visited, int curr) {
  if (curr == m.grid.length - 1) {
   visited[curr] = true; 
   path.add(curr);
   return;
  }
  path.add(curr);
  visited[curr] = true;
  int cell = m.grid[curr];
  if ((cell & Maze.LEFT) == 0 && (curr - 1) >= 0 && !visited[curr - 1] && !visited[m.grid.length - 1] && !visited[m.grid.length - 1])
   dfs(path, visited, curr - 1);
  if ((cell & Maze.RIGHT) == 0 && (curr + 1) < m.grid.length && !visited[curr + 1] && !visited[m.grid.length - 1])
   dfs(path, visited, curr + 1);
  if ((cell & Maze.UP) == 0 && (curr - m.columns) >= 0 && !visited[curr - m.columns] && !visited[m.grid.length - 1])
   dfs(path,visited, curr - m.columns);
  if ((cell & Maze.DOWN) == 0 && (curr + m.columns) < m.grid.length && !visited[curr + m.columns] && !visited[m.grid.length - 1])
   dfs(path, visited, curr + m.columns);
  if (visited[m.grid.length - 1])
   return;
  path.remove(path.size() - 1);
 }
 //using a stack
 public void solveByDFS2() {
  Stack stack = new Stack ();
  boolean[] visited = new boolean[m.grid.length];
  int[] distTo = new int[m.grid.length];
  int[] predecessor = new int[m.grid.length];
  Arrays.fill(distTo, Integer.MAX_VALUE);
  stack.push(0);
  visited[0] = true;
  distTo[0] = 0;
  predecessor[0] = -1;
  while (!stack.isEmpty() && !visited[m.grid.length - 1]) {
   int curr = stack.pop();
   int cell = m.grid[curr];
   if (curr == m.grid.length - 1) {
    break;
   }
   if ((cell & Maze.LEFT) == 0 && (curr - 1) >= 0 && !visited[curr - 1]) {
    stack.push(curr - 1);
    visited[curr - 1] = true;
    distTo[curr - 1] = distTo[curr] + 1;
    predecessor[curr - 1] = curr;
   }
   if ((cell & Maze.RIGHT) == 0 && (curr + 1) < m.grid.length && !visited[curr + 1]) {
    stack.push(curr + 1);
    visited[curr + 1] = true;
    distTo[curr + 1] = distTo[curr] + 1;
    predecessor[curr + 1] = curr;
   }
   if ((cell & Maze.UP) == 0 && (curr - m.columns) >= 0 && !visited[curr - m.columns]) {
    stack.push(curr - m.columns);
    visited[curr - m.columns] = true;
    distTo[curr - m.columns] = distTo[curr] + 1;
    predecessor[curr - m.columns] = curr;
   }
   if ((cell & Maze.DOWN) == 0 && (curr + m.columns) < m.grid.length && !visited[curr + m.columns]) {
    stack.push(curr + m.columns);
    visited[curr + m.columns] = true;
    distTo[curr + m.columns] = distTo[curr] + 1;
    predecessor[curr + m.columns] = curr;
   }
  }
  int x;
  for (x = m.grid.length - 1; distTo[x] != 0; x = predecessor[x]) {
   path.add(x);
  }
  path.add(0);
  Collections.reverse(path);
  
 }
 
 public void solveByBFS() {
  Queue q = new LinkedList ();
  boolean[] visited = new boolean[m.grid.length];
  int[] distTo = new int[m.grid.length];
  int[] predecessor = new int[m.grid.length];
  Arrays.fill(distTo, Integer.MAX_VALUE);
  q.offer(0);
  visited[0] = true;
  distTo[0] = 0;
  predecessor[0] = -1;
  while (!q.isEmpty() && !visited[m.grid.length - 1]) {
   int curr = q.poll();
   int cell = m.grid[curr];
   if (curr == m.grid.length - 1) {
    break;
   }
   if ((cell & Maze.LEFT) == 0 && (curr - 1) >= 0 && !visited[curr - 1]) {
    q.offer(curr - 1);
    visited[curr - 1] = true;
    distTo[curr - 1] = distTo[curr] + 1;
    predecessor[curr - 1] = curr;
   }
   if ((cell & Maze.RIGHT) == 0 && (curr + 1) < m.grid.length && !visited[curr + 1]) {
    q.offer(curr + 1);
    visited[curr + 1] = true;
    distTo[curr + 1] = distTo[curr] + 1;
    predecessor[curr + 1] = curr;
   }
   if ((cell & Maze.UP) == 0 && (curr - m.columns) >= 0 && !visited[curr - m.columns]) {
    q.offer(curr - m.columns);
    visited[curr - m.columns] = true;
    distTo[curr - m.columns] = distTo[curr] + 1;
    predecessor[curr - m.columns] = curr;
   }
   if ((cell & Maze.DOWN) == 0 && (curr + m.columns) < m.grid.length && !visited[curr + m.columns]) {
    q.offer(curr + m.columns);
    visited[curr + m.columns] = true;
    distTo[curr + m.columns] = distTo[curr] + 1;
    predecessor[curr + m.columns] = curr;
   }
  }
  int x;
  for (x = m.grid.length - 1; distTo[x] != 0; x = predecessor[x]) {
   path.add(x);
  }
  path.add(0);
  Collections.reverse(path); 
 }
}

Performance

Average memory usage on a 50 by 50 maze

Average running time on a 50 by 50 maze

As we can see, DFS by recursion outperforms the other two methods in both time and memory. It is understandable that BFS is slower in this circumstance. If we search layer by layer, we probably need to traverse the whole maze to find the end point. Yet if we use DFS, any path leads to the end point can terminate the searching. 

Recursive solutions often result in smaller code, which means it's more likely that the code will fit into the CPU cache. A no-recursive solution that requires an explicitly managed stack can result in larger code, more cache misses, and slower performance than recursive solution.